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Prove that $\dfrac{1}{\dfrac{1}{a}+\dfrac{1}{b}}+\dfrac{1}{ \dfrac{1}{c}+\dfrac{1}{d}} \le \dfrac{1}{\dfrac{1}{a+c}+\dfrac{1}{b+d}}$ for all positive real numbers $a, b, c, d$.
Hello.Prove that $\dfrac{1}{\dfrac{1}{a}+\dfrac{1}{b}}+\dfrac{1}{ \dfrac{1}{c}+\dfrac{1}{d}} \le \dfrac{1}{\dfrac{1}{a+c}+\dfrac{1}{b+d}}$ for all positive real numbers $a, b, c, d$.
Thanks for your post, mente oscura! I have been trying to digest what you have come up with for some time, hoping to see the light at the end of the tunnel but I am sorry, I couldn't possibly continue from there.Hello.
I can't think of another thing that solved with the limits.
Making calculations.
[tex]T=\dfrac{1}{1+\frac{d}{b}+\frac{b}{a}+\frac{d}{a}}+ \dfrac{1}{1+\frac{b}{d}+\frac{b}{c}+\frac{d}{c}}+ \dfrac{1}{1+\frac{a}{b}+\frac{c}{b}+\frac{c}{a}}+ \dfrac{1}{1+\frac{a}{d}+\frac{a}{c}+\frac{c}{d}} \le{1}[/tex]
We note that, when they are the same two by two, the result is 1.
Now let's look:
[tex]\displaystyle\lim_{a \to{+}\infty}{T}=\dfrac{1}{1+\frac{d}{b}}+ \dfrac{1}{1+\frac{b}{d}+\frac{b}{c}+\frac{d}{c}}[/tex]
[tex]\displaystyle\lim_{b \to{+}\infty}{} (\dfrac{1}{1+\frac{d}{b}}+ \dfrac{1}{1+\frac{b}{d}+\frac{b}{c}+\frac{d}{c}})=1[/tex]
It would be like if we choose any of the other variables, because "T" combinations are symmetric.
Regards.
Hello. Uff.Thanks for your post, mente oscura! I have been trying to digest what you have come up with for some time, hoping to see the light at the end of the tunnel but I am sorry, I couldn't possibly continue from there.
I want to share the solution that I found with you and the rest of the members:
If we shift everything from the left to the right, we get
$\dfrac{1}{\dfrac{1}{a}+\dfrac{1}{b}}+\dfrac{1}{ \dfrac{1}{c}+\dfrac{1}{d}} \le \dfrac{1}{\dfrac{1}{a+c}+\dfrac{1}{b+d}}$
...
Prove that $\dfrac{1}{\dfrac{1}{a}+\dfrac{1}{b}}+\dfrac{1}{ \dfrac{1}{c}+\dfrac{1}{d}} \le \dfrac{1}{\dfrac{1}{a+c}+\dfrac{1}{b+d}}$ for all positive real numbers $a, b, c, d$.
may be I am missing something$\dfrac{1}{\dfrac{1}{a}+\dfrac{1}{b}}+\dfrac{1}{ \dfrac{1}{c}+\dfrac{1}{d}}---
(1)$
$ \dfrac{1}{\dfrac{1}{a+c}+\dfrac{1}{b+d}}----(2)$
using $AP\geq GP :$
(1)$\leq\dfrac{\sqrt {ab}}{2}+\leq\dfrac{\sqrt {cd}}{2} $---(3)
(2)$\leq\dfrac{\sqrt {(a+c)(b+d)}}{2}$---(4)
$(3)^2=\dfrac {ab+cd+2\sqrt {abcd}}{4} ----(5)$
$(4)^2=\dfrac {ab+ad+bc+bd}{4}----(6)$
again using $AP\geq GP :$
$(6)\geq (5)$
and the proof is done
sory a typo in (6)
$(3)^2=\dfrac {ab+cd+2\sqrt {abcd}}{4} ----(5)$
$(4)^2=\dfrac {ab+ad+bc+cd}{4}----(6)$
$ad+bc\geq 2\sqrt {abcd}$
(6) > (5) => (3) < (4) is true
But (3) < (4) and (1) < (3) and (2) < (4) does not mean (1) < (2) .